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# sm8_44

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```Problem 8.44 The arrival times of two signals at a receiver are uniformly distributed over the interval
[0,T]. The receiver will be jammed if the time difference in the arrivals is less than \u3c4. Find the probability
that the receiver will be jammed.

Solution
Let X and Y be random variables representing the arrival times of the two signals. The
probability density functions of the random variables are

( )
\u23aa\u23a9
\u23aa\u23a8
\u23a7 <<=
otherwise,0
01 Tx
Txf X

and fY(y) is similarly defined. Then the probability that the time difference between
arrivals is less than \u3c4 is given by
[ ] [ ] [ ] [ ] [ ]
[ ]YXYX
XYXYYXYXYXYXYX
><\u2212=
>><\u2212+>><\u2212=<\u2212
|
||
\u3c4
\u3c4\u3c4\u3c4
P
PPPPP

where the second line follows from the symmetry between the random variables X and Y,
namely, P[X > Y] = P[Y > X]. If we only consider the case X > Y, then we have the
conditions: 0 < X < T and 0 < Y < X < \u3c4+Y. Combining these conditions we have
Y < X < min(T, \u3c4+Y). Consequently,

[ ] ( )
( )
( ){ }\u222b
\u222b \u222b
\u222b \u222b
\u2212+=
\u239f\u23a0
\u239e\u239c\u239d
\u239b=
=<\u2212
+
+
T
T yT
y
T yT
y
YX
dyyyT
T
dydx
T
dydxyfxfYX
0
2
0
,min 2
0
,min
,min1
1
)()(
\u3c4
\u3c4
\u3c4
\u3c4
P

Combining the two terms of the integrand,

[ ] ( )
\u239f\u23a0
\u239e\u239c\u239d
\u239b=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212=
\u2212=<\u2212 \u222b
T
yyTy
T
dyyT
T
YXP
T
T
\u3c4
\u3c4
\u3c4\u3c4
,
2
1min
,
2
min1
,min1
0
2
2
0
2

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page\u20268-54```